Suppose a laboratory has a 31 g sample of polonium-210. The half-life of polonium-210 is about 138 days. How many half lives of polonium occur in 966 days? How much polonium is in the sample 966 days later?

1 Answer
May 19, 2017

Answer:

#7# half-lives

#0.24g# #"^(210)Po#

Explanation:

The number of half-lives is simply the number of days given (#966d#) divided by the half-life (#138d#), which is #7# half-lives.

The amount of #"^(210)Po# remaining after #t# days is represented by the equation

#N(t) = N_0(1/2)^(t/(t_(1/2))#

where #N(t)# is the amount of substance remaining after #t# days,
#N_0# is the initial amount of substance,
#t# is the time in days, and
#t_(1/2)# is the half-life of the substance.

Plugging in known variables, we have

#N(t) = (31g)(1/2)^((966)/(138)) = 0.24g# #"_(210)Po#